How many points of intersection between an ellipse and an $L_p$-circle? Consider an ellipse $E$ in the plane, centered at the origin.  (In my case, the minor axis points into the nonnegative quadrant.)
Let S be an "$L_p$-circle":  $S = \{(x,y) : |x|^p + |y|^p = 1\}$, where $p > 1$, $p \neq 2$.
Is it correct that $E$ and $S$ have at most 8 intersections?
More exactly what I care about: is it true that $E$ and $S$ have at most 2 intersections in the nonnegative quadrant?
It seems relatively clear from drawing pictures, but I couldn't quite get the natural convexity argument to work. 
 A: This question has been asked and answered on MathOverflow. I have replicated the accepted answer by Ilya Bogdanov below.

$\def\sign{\mathop{\rm sign}}$First of all, it is enough to prove the statement when $p=u/v$ is rational, $u$ is even and $v$ is odd (such numbers are dense on the real line). We need this to simplify the last argument.
Let the equation of the ellipse be $f(x,y)=ax^2+2bxy+cy^2=1$; one may assume that $b\neq 0$ (either by a small variation argument, or by considering it directly --- this case is easy). Let us bound the number of local extrema of $f(x,y)$ on $|x|^p+|y|^p=1$; if there are at most 8 of them, then we are done: between every two, there is at most one intersection point. Since $b\neq 0$, the points where one of the coordinates vanish are not extremal (the ellipse passing through such point has a tangent not parallel to the coordinate axes).
The extremal points satisfy the system of Lagrange equations
  $$
  2(ax+by)=p\lambda|x|^{p-1}\sign x, \quad
  2(bx+cy)=p\lambda|y|^{p-1}\sign y;
$$
  obviously, $\lambda\neq 0$, so this system yields 
  $$
  \frac{at+b}{bt+c}=|t|^{p-1}\sign t,
$$
  where $t=x/y$. So, it suffices to show that there are at most four such values of $t$ (each corresponds to two symmetrical extremal points). Notice that $|t|^{p-1}\sign t=t^{(u-v)/v}$ by our assumptions on $u$ and $v$. Now, substituting $t=s^v$ we obtain
  $$
  as^v+b=bs^u+cs^{u-v}.
$$
  The last polynomial equation has at most four roots by an easy application of Descartes' rule of signs.
ADDENDUM. By the way, you have mentioned on MathSE that in your case the minor axis is in the positive quadrant, and you were interested in the intersection points lying in this quadrant as well. In this case we may assume that $a,b,c>0$, and we are interested in positive values of $t$. The rule of signs shows that there is at most one $t$ whenever $u-v>v$, that is --- if $p>2$. Otherwise it claims that there is at most one $t$ in the adjacent quadrant.
ADDENDUM2. Here is the explanation about the limiting case. Assume that you have more than 8 common pointsof an ellipse and an $\ell_p$-circle. Blowing up or away your ellipse a bit, you may reach the situation when there are more than 8 points of transversal intersection. Then you mark a point on the ellipse between every such neighboring points (or simply mark 10 of them); these marked points are alternately inside and outside the $\ell_p$-circle. Finally, if you change $p$ a bit, all these 10 points will still be alternating, guaranteeing 10 intersection points.
The same method works for $b=0$.
Finally, if it is easier for you, you may pass to rational $p$ of a desired form AFTER writing down the (almost) polynomial in $t$ --- in this case the statement is more visible...
And, even more finally, you may just repeat the proof of the Descartes' rule for an `almost polynomial' --- all you (perhaps) need is that all the powers differ by at least one, which can be reached by an appropriate substitution.
ADDENDUM3. Just to mention. The same method allows to prove that the affine images of $\ell_p$- and $\ell_q$-circles also have at most 8 common points, if they have the same center.

